Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <iomanip>
using namespace std;
int N;
vector<double> p;
double dp[3100][3100];
int main(){
	cin>>N;
	p.resize(N);
	for (int i=0;i<N;++i)cin>>p[i];
	dp[0][0]=1.0;
	for (int i=0;i<N;++i){
		for (int j=0;j<=i;++j){
			dp[i+1][j+1]+=dp[i][j]*p[i];
			dp[i+1][j]+=dp[i][j]*(1.0-p[i]);
		}
	}

	double res=0.0;
	for (int j=N/2+1;j<=N;++j)res+=dp[N][j];
	cout<<fixed<<setprecision(10)<<res<<endl;

}